Question

A block of mass m= 4 kg is placed over a rough inclined plane as shown in Fig. The coefficient of friction between the block and the plane is $\mu =0.5$. A force F = 10N is applied on the block at an angle of ${30}^0$. Find the contact force between the block and the plane.

• A. $10.5\sqrt{5}$ N
• B. $13.5\sqrt{5}$ N
• C. $12.5\sqrt{5}$ N
• D. $10\sqrt{5}$ N

Answer 1

The correct option is B $13.5\sqrt{5}$ N

$\begin{array}{c}Sincethemassisbalanced,{\sum }^{}{F}_x=0,\sum F_y=0\\ mg\sin {37}^0-F\cos {30}^0+F_x=0\\ 40\times \frac{3}{5}-10\times \frac{\sqrt{3}}{2}+F_x=0\\ 24-5\sqrt{3}+F_x=0\\ F_x=-24+3.66\\ =-15.34\\ or,\\ mg\cos {37}^0-F\sin {30}^0+N_y=0\\ 40\times \frac{4}{5}-10\times \frac{1}{2}+N_y=0\\ 32-5+N_y=0\\ -27=N_y\\ Now,\\ Totalcontactforce=\sqrt{F_x^2+N_y^2}=13.5\sqrt{5}\\ Hence,\\ optionBiscorrectanswer.\end{array}$